Class 7, Maths, Chapter 10, Exercise 10.3 Solutions
Q.1. Construct ∆DEF such that DE = 5 cm, DF = 3 cm and m∠EDF = 90°.
Ans:
Steps of construction:
(a) Draw a line segment DF = 3 cm.
(b) At point D, draw a ray DX to making an angle of 90o i.e. ∠XDF = 90o.
(c) Taking D as centre, draw an arc of radius 5 cm, which cuts DX at the point E.
(d ) Join EF.
Then, It is the required right angled triangle ΔDEF.
Q.2. Construct an isosceles triangle in which the lengths of each of its equal sides is 6.5 cm and the angle between them is 110°.
Ans:
Steps of construction:
- Draw a line segment AB = 6.5 cm.
- At point A, draw a ray AX to making an angle of 110oi.e. ∠XAB = 110o.
- Taking A as centre, draw an arc with radius 6.5 cm, which cuts AX at point C.
- Join CB.
Then, ΔABC is the required isosceles triangle.
Q.3. Construct ∆ABC with BC = 7.5 cm, AC = 5 cm and m∠C = 60°.
Ans:
Steps of construction:
- Draw a line segment BC = 7.5 cm.
- At point C, draw an angle of 60° with the help of protractor, i.e. ∠XCB = 60o.
- Taking C as centre and radius 5 cm, draw an arc, which cuts XC at point A.
- Join AB.
Then, ΔABC is the required triangle.
NCERT Solutions For Class 7 Maths, Chapter 10, Practical Geometry (All Exercises)
Class 7, Maths, Chapter 9, Practical Geometry
Class 7, Maths, Chapter 9, Practical Geometry, Exercise 10.1
Class 7, Maths, Chapter 9, Practical Geometry, Exercise 10.2
Class 7, Maths, Chapter 9, Practical Geometry, Exercise 10.3 ← You are here
Class 7, Maths, Chapter 9, Practical Geometry, Exercise 10.4
Class 7, Maths, Chapter 9, Practical Geometry, Exercise 10.5